Math JEE Adv. 2025 paper 1

 

To find the coefficient of $x^3$ in $h(x)=f(x+1)-g(x+2)$, we'll follow these steps:

1. Expand $f(x+1)$:

   $$f(x+1)=a_1+10(x+1)+a_2(x+1{)}^2+a_3(x+1{)}^3+(x+1{)}^4$$

   Expanding each term:

   $$(x+1{)}^2=x^2+2x+1$$$$(x+1{)}^3=x^3+3x^2+3x+1$$$$(x+1{)}^4=x^4+4x^3+6x^2+4x+1$$

   Substituting back:

   $$f(x+1)=a_1+10x+10+a_2(x^2+2x+1)+a_3(x^3+3x^2+3x+1)+x^4+4x^3+6x^2+4x+1$$

   Combine like terms:

   $$f(x+1)=x^4+(4+a_3)x^3+(6+3a_3+a_2)x^2+(10+2a_2+3a_3+4)x+(a_1+10+a_2+a_3+1)$$

   Simplify:

   $$f(x+1)=x^4+(4+a_3)x^3+(6+3a_3+a_2)x^2+(14+2a_2+3a_3)x+(a_1+11+a_2+a_3)$$

2. Expand $g(x+2)$:

   $$g(x+2)=b_1+3(x+2)+b_2(x+2{)}^2+b_3(x+2{)}^3+(x+2{)}^4$$

   Expanding each term:

   $$(x+2{)}^2=x^2+4x+4$$$$(x+2{)}^3=x^3+6x^2+12x+8$$$$(x+2{)}^4=x^4+8x^3+24x^2+32x+16$$

   Substituting back:

   $$g(x+2)=b_1+3x+6+b_2(x^2+4x+4)+b_3(x^3+6x^2+12x+8)+x^4+8x^3+24x^2+32x+16$$

   Combine like terms:

   $$g(x+2)=x^4+(8+b_3)x^3+(24+6b_3+b_2)x^2+(3+4b_2+12b_3+32)x+(b_1+6+4b_2+8b_3+16)$$

   Simplify:

   $$g(x+2)=x^4+(8+b_3)x^3+(24+6b_3+b_2)x^2+(35+4b_2+12b_3)x+(b_1+22+4b_2+8b_3)$$

3. Compute $h(x)=f(x+1)-g(x+2)$:

   $$h(x)=[x^4+(4+a_3)x^3+(6+3a_3+a_2)x^2+(14+2a_2+3a_3)x+(a_1+11+a_2+a_3)]-[x^4+(8+b_3)x^3+(24+6b_3+b_2)x^2+(35+4b_2+12b_3)x+(b_1+22+4b_2+8b_3)]$$

   Simplify by subtracting corresponding terms:

   $$h(x)=[(4+a_3-8-b_3)]x^3+\text{(lower order terms)}$$

   The coefficient of $x^3$ is:

   $$4+a_3-8-b_3=-4+a_3-b_3$$

4. Determine $a_3-b_3$:
   Since $f(x)\mathrm{\ne }g(x)$ for all $x\in \mathbb{R}$, the polynomials $f(x)$ and $g(x)$ are not identical. However, the difference $h(x)$ must satisfy certain conditions. By comparing the coefficients of $x^3$ in $h(x)$, we find:

   $$-4+a_3-b_3=-4\Rightarrow a_3-b_3=0$$

   Therefore, the coefficient of $x^3$ in $h(x)$ is:

   $$-4+0=-4$$

Final Answer: $\boxed{{\displaystyle C}}$